Rotation 4 Question 4

5. A thin ring of mass 2kg and radius 0.5m is rolling without slipping on a horizontal plane with velocity 1m/s. A small ball of mass 0.1kg, moving with velocity 20m/s in the opposite direction, hits the ring at a height of 0.75m and goes vertically up with velocity 10m/s. Immediately after the collision,

(2011)

(a) the ring has pure rotation about its stationary CM

(b) the ring comes to a complete stop

(c) friction between the ring and the ground is to the left

(d) there is no friction between the ring and the ground

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Answer:

Correct Answer: 5. (a,c)

Solution:

  1. The data is incomplete. Let us assume that friction from ground on ring is not impulsive during impact.

From linear momentum conservation in horizontal direction, we have

(2×1)+(0.1×20)ve=(0.1×0)+(2×v)+ve

Here, v is the velocity of CM of ring after impact. Solving the above equation, we have v=0

Thus, CM becomes stationary.

Correct option is (a).

Linear impulse during impact

(i) In horizontal direction

J1=Δp=0.1×20=2Ns

(ii) In vertical direction J2=Δp=0.1×10=1N-s Writing the equation (about CM )

Angular impulse = Change in angular momentum

We have,

1×32×122×0.5×12=2×(0.5)2ω10.5

Solving this equation ω comes out to be positive or ω anti-clockwise. So just after collision rightwards slipping is taking place.

Hence, friction is leftwards.

Therefore, option (c) is also correct.



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